Question

Use the binomial formula for the first three terms of $(1+x)^n$ for negative and fractional indices $(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2$ To find the first three terms of the expansion of $(1+x)^{0.25}$ Enter the coefficients corrected (if necessary) to three decimal places. e.g. $2 + 3x + 3.667x^2$

          Use the binomial formula for the first three terms of $(1+x)^n$ for negative and fractional indices
$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2$
To find the first three terms of the expansion of
$(1+x)^{0.25}$
Enter the coefficients corrected (if necessary) to three decimal places.
e.g. $2 + 3x + 3.667x^2$

$Use the binomial formula for the first three terms of (1+x)^n for negative and fractional indices (1+x)^n = 1 + nx + (n(n-1))/(2!)x^2 To find the first three terms of the expansion of (1+x)^0.25 Enter the coefficients corrected (if necessary) to three decimal places. e.g. 2 + 3x + 3.667x^2$

Added by Joseph U.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Sri K

02/26/2023

Step 1

Step 1: The binomial formula for (1+x)^(n) is given by: (1+x)^(n) = 1 + nx + (n(n-1))/(2!)x^(2) Show more…

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Use the binomial formula for the first three terms of (1+x)^(n) for negative and fractional indices. (1+x)^(n) = 1 + nx + (n(n-1))/(2!)x^(2) To find the first three terms of the expansion of (1+x)^(0.25), enter the coefficients corrected (if necessary) to three decimal places. 6.9 + 3x + 3.667x^(2) Use the binomial formula for the first three terms of 1+x for negative and fractional indices. (1+x) = 1 + nx To find the first three terms of the expansion of (1+x)^0.25, enter the coefficients corrected (if necessary) to three decimal places. 2 + 3x + 3.667x^2